Mg + 2HCl -> MgCl2 +H2 delta H = ?
MgO + 2HCl (aq) -> MgCl2+ H2O delta H = ?
H2+ 1/2O2 -> H2O delta H = -285.8 kJ
Using Hess' Law find both the missing delta H values.
Mg + 2HCl -> MgCl2 +H2: delta H = -571.6 kJ
MgO + 2HCl (aq) -> MgCl2+ H2O: delta H = -857.4 kJ
To find the missing ΔH values using Hess's Law, we need to manipulate and combine the given equations to get the desired reaction.
Let's start with reaction 1:
Mg + 2HCl → MgCl2 + H2 ΔH = ?
To find the enthalpy change for this reaction, we will consider reaction 2:
MgO + 2HCl → MgCl2 + H2O ΔH = ?
We can see that reaction 2 is a combination of reaction 1 and an additional step involving the formation of MgO.
Step 1: Reverse reaction 1:
MgCl2 + H2 → Mg + 2HCl ΔH = -ΔH1 (Note that the sign of ΔH is reversed when we reverse the reaction.)
Step 2: Multiply reaction 1 by 1/2 to balance the number of moles of HCl:
1/2(Mg + 2HCl → MgCl2 + H2) ΔH = 1/2 * ΔH1
Step 3: Add the equations:
1/2(Mg + 2HCl → MgCl2 + H2) + (MgO + 2HCl → MgCl2 + H2O) ΔH = 1/2 * ΔH1 + ΔH2
Now, let's consider reaction 3:
H2 + 1/2O2 → H2O ΔH = -285.8 kJ
Step 4: Multiply reaction 3 by 2 to balance the number of moles of H2:
2(H2 + 1/2O2 → H2O) ΔH = -2 * 285.8 kJ
Since we want reaction 2 to match reaction 3, we will reverse reaction 2 and multiply it by 2:
2(MgCl2 + H2O → MgO + 2HCl) ΔH = -2 * ΔH2
So, our final equation to find the desired reaction is:
1/2(Mg + 2HCl → MgCl2 + H2) + 2(MgCl2 + H2O → MgO + 2HCl) = 2(H2 + 1/2O2 → H2O)
Now, we can use the equation above to find the missing ΔH values:
1/2 * ΔH1 + ΔH2 = -2 * 285.8 kJ
Simplifying the equation, we have:
ΔH1 + 2ΔH2 = -571.6 kJ
Using the values given in the question, you can solve for ΔH1 and ΔH2 by plugging in the appropriate values and then solve the resulting equation.
Hess' Law is a principle in thermodynamics that states that the overall enthalpy change for a reaction is the sum of the enthalpy changes of individual reactions that lead to the same final products.
To find the missing ΔH values using Hess' Law, we need to use a combination of known reactions to derive the desired reaction. Let's go step-by-step:
Step 1: Write the known reactions:
Reaction 1: Mg + 2HCl -> MgCl2 + H2 ΔH = ?
Reaction 2: MgO + 2HCl -> MgCl2 + H2O ΔH = ?
Step 2: Identify any common compounds in both reactions. In this case, both reactions have MgCl2.
Step 3: Manipulate the reactions to match the desired reaction:
In Reaction 1, we have Mg and HCl on the reactant side, while in Reaction 2 we have MgO and HCl. To get the desired reaction, we need to cancel Mg and HCl from one of the reactions and match the other reactants and products.
Step 4: Manipulate Reaction 1:
Multiply Reaction 1 by 1/2 to balance the number of moles of HCl and H2:
1/2(Mg + 2HCl -> MgCl2 + H2) ΔH = 1/2ΔH1
Step 5: Manipulate Reaction 2:
Multiply Reaction 2 by 1 to balance the number of moles of HCl:
1(MgO + 2HCl -> MgCl2 + H2O) ΔH = ΔH2
Step 6: Combine the manipulated reactions:
Adding Reaction 1 and Reaction 2 gives us the desired reaction:
1/2(Mg + 2HCl -> MgCl2 + H2) + 1(MgO + 2HCl -> MgCl2 + H2O) = 1/2(Mg + MgO + 4HCl -> 2MgCl2 + H2 + H2O)
Step 7: Simplify the combined reaction:
1/2(Mg + MgO + 4HCl -> 2MgCl2 + H2 + H2O) = 1/2(Mg + MgO + 4HCl -> 2MgCl2 + 2H2O) ΔH = 1/2ΔH1 + ΔH2
The ΔH for the overall reaction can be expressed as the sum of the enthalpy changes for the individual reactions, so:
ΔH = 1/2ΔH1 + ΔH2
By applying this principle, we can calculate the missing ΔH values.